Let k be the radius of inversion, so that Oɑ.OA

EXERCISES CXXVII.

EXAMPLES ON INVERSION.

1. Invert the following theorems :

(a) The sum of the three angles of any triangle is equal to two right angles.

(b) AB and CD are two parallel lines. A line which is drawn at right angles to AB, is also at right angles to CD.

(c) If a transversal meets two other lines in the same plane, making the two alternate angles equal to one another, these two straight lines shall be parallel. (Centre of inversion on transversal.)

2. Two given circles touch at p, and O is a point lying on neither. Another circle through O and p meets the two circles again in R, r. S, s are corresponding points for a second circle through O and p. Prove that the circumcentres of Ors, ORS are collinear with 0.

3. A straight line touches a circle at A, and 0 is another point outside them. A circle is drawn through O touching the line in P and the circle in Q. Prove that the circumcircles of OAP, OAQ, cut orthogonally.

4. Two given circles touch at A, and O is any point outside them. Any circle through 0 and A intersects the given circles in B and C. A circle is drawn through O touching at B the circle AB, and another circle touching at C the circle AC. Prove that these two circles touch one another at 0.

5. Three circles pass through a point 0. Their other intersections determine a curvilinear triangle ABC. P is any point on the circumcircle of this curvilinear triangle. Circles are drawn through P and O orthogonally to the three circles, and meet them again in D, E, F. Prove that ODEF is cyclic.

CHAPTER XXV.

MISCELLANEOUS THEOREMS IN HIGHER

GEOMETRY.

THE RADICAL AXIS.

350. Recapitulation. The Definitions, Theorems and Problems in this article have already been given in § 256, Chapter XVI.

DEFINITION.

The radical axis of two given circles is the locus of a point from which the tangents to the two circles are of equal length.

THEOREM 1.—The radical axis of two circles is a straight line perpendicular to the join of their centres.

PROBLEM 1. To construct the radical axis of two given

circles.

[Cons.-Draw any circle to cut one of the given circles at G, H and the other at K, L. Through the intersection of GH, KL draw a line perpendicular to the join of the centres of the given circle.]

THEOREM 2. Given three circles X, Y, Z, the radical axes of the three pairs of circles (X, Y), (Y, Z), (Z, X) are concurrent. DEFINITION.-The radical centre of three circles is the point of intersection of their three radical axes.

351. THEOREM 3. The difference between the squares on the tangents from a given point to two given circles is double of the rectangle contained by the join of the centres and the distance of the given point from the radical axis.

Given that XY is the radical axis (Fig. 488).
Required to prove that PC2 - PD2 = 2PN.AB.
Cons.-Draw PL LAB, and bisect AB at M.

FIG. 488.

In A APB, PLLAB and AB is bisected at M, hence (by T.10) PA2-PB22AB.ML.

.

X is on the radical axis of the two circles, .. (by § 256), AX2 - XB2 = AC2 - BD2 Also

AX2 - XB2 = (AX + XB)(AX - XB)

= AB.2MX

AC2 BD2 2AB.MX

AC2 - BD2

(v), that

=

2AB.MX

2AB.ML

=

Thus

DEFINITION.-A system of circles is coaxial, if the radical axis of any two circles in the system is always the same line.

Thus, if in Fig. 489 the line XY is the radical axis of any pair of circles, then the system is coaxial.

THEOREM 4.-C1, C2, C3,

=

are a system of circles whose radii are 71, 72, 73, and whose centres 01, 02, 03, .. all lie on a line AXB which is perpendicular to the line XY. Then (i) if 01X2 - rı2 = 02X2 -- r22 the circles are coaxial, and XY is their common radical axis; (ii) conversely, if the circles are coaxial, and XY is their common radical axis, then 01X2 - r12 = 02X2 − r22 = 03X2 – 1°32 =

=

Let P be any point on XY, and let PT1, PT1⁄2, be tangents from P to the circles.

Required to prove that XY is the radical axis of any

pair of circles in the system.

=

22.

229

2

PT2, i. e. PT1 = PT2.

(Hyp.)

Similarly the tangents to the circles C1, C2 from any other point on XY are equal.

Hence XY is the radical axis of the circles C1, C2

Similarly XY is the radical axis of any other pair. (ii) It is obvious that the converse theorem can be proved by reversing the argument in (i).

SIM. GEO.

I I

353.* Limiting points and common points.-There are three cases in Theorem 4, according as the quantities OX2 – r ̧2, О„X2 – r‚22, 03×2 – r2, etc., are all positive, all negative or all zero.

CASE I.-Let 01X2 – r12 = 02X2 – r‚2 : In this case r1<01X, ï1⁄2<02X, sects the radical axis (Fig. 489).

= + 12.

. . Hence none of the circles inter

It follows that no two of the circles intersect one another; for if they did, their radical axis would be their common chord, and would therefore intersect them (see Exercises XCVI, 2).

=

=

Again in Fig. 489 let XH XKl. Then the circle with centre H and radius zero is a circle of the coaxial system; for it satisfies the conditions of Theorem 4. [To verify this prove_PH = PT1 = . . .] Similarly for the circle with the centre K, and radius zero. The "point-circles" H, K are called the limiting points of the system of coaxial circles.

Again, in Fig. 490, let XF = XG = 1.
Then

=

12.

Hence all the circles intersect

01X2 + 12 = 01X2 + XF2 = 01F2 = 01G2.

Hence circle C1 passes through Ƒ and G.

Similarly every circle of the system passes through F and G. Thus F and G are the common points of this system of coaxial circles. There are no limiting points (i. e. "point-circles" belonging to the system), for the radius cannot be less than 1.

CASE III.-Let O1X2 - r12 = 02X2 – r22 :

=0.

In this case it is obvious that all the circles touch the radical axis at X. The two limiting points of the system coincide with X.